In this HackerEarth Distinct Count problem solution we have given an array A of N integers, classify it as being Good Bad, or Average. It is called Good, if it contains exactly X distinct integers, Bad if it contains less than X distinct integers, and Average if it contains more than X distinct integers.


HackerEarth Distinct Count problem solution


HackerEarth Distinct Count problem solution.

#include <bits/stdc++.h>
using namespace std;
int main () {
int tc;
scanf("%d",&tc);
while (tc--) {
int n, k;
scanf("%d%d",&n,&k);
set < int > s;
int temp;
for (int i=0; i<n; i++) {
scanf("%d",&temp);
s.insert(temp);
}

if (s.size()<k) {
printf("Bad husband\n");
}
if (s.size()==k) {
printf("Perfect husband\n");
}
if (s.size()>k) {
printf("Lame husband\n");
}
}
return 0;
}

Second solution

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define vi vector<int>
#define vl vector<ll>
#define pii pair<int,int>
#define pil pair<int, ll>
#define pll pair<ll, ll>
#define pli pair<ll, int>
#define pb(v, a) v.push_back(a)
#define mp(a, b) make_pair(a, b)
#define MOD 1000000007
#define rep(i, a, b) for(i=a; i<=b; ++i)
#define rrep(i, a, b) for(i=a; i>=b; --i)
#define si(a) scanf("%d", &a)
#define sl(a) scanf("%lld", &a)
#define pi(a) printf("%d", a)
#define pl(a) printf("%lld", a)
#define pn printf("\n")
ll pow_mod(ll a, ll b)
{
ll res = 1;
while(b)
{
if(b & 1)
res = (res * a) % MOD;
a = (a * a) % MOD;
b >>= 1;
}
return res;
}
set<int> s;
int main()
{
int t, i, n, x, j, tmp;
si(t);
assert(t >= 1 && t <= 50);
rep(i, 1, t)
{
s.clear();
si(n);
si(x);
assert(n >= 1 && n <= 13000);
assert(x >= 1 && x <= 13000);
rep(j, 1, n)
{
si(tmp);
assert(tmp >= 1 && tmp <= 1000000000);
s.insert(tmp);
}
if(s.size() < x)
printf("Bad husband\n");
else if(s.size() > x)
printf("Lame husband\n");
else
printf("Perfect husband\n");
}
return 0;
}